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Rashid [163]
3 years ago
11

If a pendulum swings with a period of 1.00 s and has a length of 0.250 m, calculate the magnitude of the acceleration due to gra

vity.
Physics
1 answer:
Delicious77 [7]3 years ago
3 0

Answer:

The magnitude of g is 9.870 m/s^2

Explanation:

Use the formula tying the period to the length of a pendulum for small amplitudesT=2\pi\sqrt{\frac{L}{g}}\implies g = \frac{4\pi^2 }{T^2}L=9.870\frac{m}{s^2}:


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Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i
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Answer:

The value is  r =  5.077 \  m

Explanation:

From the question we are told that

   The  Coulomb constant is  k =  9.0 *10^{9} \  N\cdot  m^2  /C^2

   The  charge on the electron/proton  is  e =  1.6*10^{-19} \  C

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Generally for the electron to be held up by the force gravity

   Then    

       Electric force on the electron  =  The  gravitational Force

i.e  

            m_{electron} *  g  = \frac{ k *  e^2  }{r^2 }

         \frac{9*10^9 *  (1.60 *10^{-19})^2  }{r^2 }  =     9.11 *10^{-31 }  *  9.81

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         r =  5.077  \  m

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Which object is I’m equilibrium?
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